Monotonicity of Sequences

Bounded Sequences

Definition Upper and Lower Bounds

Let $(u_n)$ be a sequence.

$(u_n)$ is upper bounded if there exists a real $M$ such that for all $n$, $\boldsymbol{u_n \le M}$. $M$ is an upper bound.
$(u_n)$ is lower bounded if there exists a real $m$ such that for all $n$, $\boldsymbol{u_n \ge m}$. $m$ is a lower bound.
$(u_n)$ is bounded if it is both upper and lower bounded.

Example

Consider $u_n = 1 + \frac{2}{n}$ for $n \ge 1$.

For $n \ge 1$, we have $$ \begin{aligned} n &\ge 1 \\ \frac{n}{n} &\ge \frac{1}{n} \\ \frac{1}{n} &\le 1 \\ \frac{2}{n} &\le 2 \\ 1+\frac{2}{n} &\le 1+2 \\ u_n &\le 3. \end{aligned} $$ Therefore, the sequence $(u_n)$ is bounded above by $3$ (i.e. $3$ is an upper bound).
For $n \ge 1$, we have$$\begin{aligned}n &\ge 1 \\ n &> 0 \\ \frac{1}{n} &> 0 \quad (\text{the reciprocal has the same sign})\\ \frac{2}{n} &> 0 \\ 1+\frac{2}{n} &> 1 \\ u_n &> 1.\end{aligned}$$Therefore, the sequence $(u_n)$ is bounded below by $1$ (i.e. $1$ is a lower bound).
Therefore, $(u_n)$ is bounded.

Proposition Unbounded Monotonic Sequences

If a sequence is increasing and not upper bounded, then it diverges to $+\infty$.
If a sequence is decreasing and not lower bounded, then it diverges to $-\infty$.

Theorem Monotone Convergence Theorem

Attention!

This theorem proves that a limit exists, but it does not necessarily give the value of the limit.

Example

Let $(u_n)$ be the sequence defined for $n \ge 0$ by $u_{n}=\dfrac{n-1}{n+4}$.

Boundedness (upper bound): $$\begin{aligned}1 - u_n &= 1 - \dfrac{n-1}{n+4}\\ 1 - u_n&= \dfrac{n+4}{n+4} - \dfrac{n-1}{n+4}\\ 1 - u_n&= \dfrac{n+4 - (n-1)}{n+4}\\ 1 - u_n&= \dfrac{5}{n+4}\\ 1 - u_n&>0\\ 1 &>u_n\\ \end{aligned}$$ For $n\ge 0$, $n+4>0$, hence $\dfrac{5}{n+4}>0$, so $1-u_n>0$, i.e. $u_n<1$. So, $(u_n)$ is upper bounded by $1$.
Monotonicity: $$ \begin{aligned} u_{n+1}-u_n &= \frac{(n+1)-1}{(n+1)+4}-\frac{n-1}{n+4}\\ &= \frac{n}{n+5}-\frac{n-1}{n+4}\\ &= \frac{n(n+4)-(n-1)(n+5)}{(n+5)(n+4)}\\ &= \frac{n^2+4n-\bigl(n^2+5n-n-5\bigr)}{(n+5)(n+4)}\\ &= \frac{n^2+4n-(n^2+4n-5)}{(n+5)(n+4)}\\ &= \frac{5}{(n+5)(n+4)}\\ &>0\\ \end{aligned} $$ So, $(u_n)$ is increasing.
Conclusion:
Since $(u_n)$ is increasing and upper bounded by $1$, the Monotone Convergence Theorem implies that $(u_n)$ converges.